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A solenoid with a diameter of 6.35 cm and a length of 20.3 cm has 186 turns and carries a current of 13.3 A. Find the magnetic flux through the circular cross-sectional area of the solenoid. Recall [tex]\mu_0 = 1.25664 \times 10^{-6} \, \text{T}\cdot\text{m/A}[/tex].

Answer :

Final answer:

The magnetic flux through the circular cross-sectional area of the solenoid is found by first calculating the magnetic field inside the solenoid and then multiplying it by the coil's cross-sectional area. This yields a magnetic flux of approximately 0.0000474 Weber.

Explanation:

To find the magnetic flux through the circular cross-sectional area of the solenoid, we first need to calculate the magnetic field inside the solenoid. The formula for the magnetic field inside a solenoid is B = µ0nI, where n represents the number of turns per meter of the solenoid and I is the electrical current (13.3 A in this case). The number of turns (n) can be found by dividing the total number of turns (186) by the length of the solenoid (20.3 cm = 0.203 m), yielding approximately 916.3 turns/meter.

Substituting in the given values, we have B = 1.25664×10-6 T*m/A * 916.3 turns/m * 13.3 A = 0.015 T (Tesla). The magnetic flux (Φ) through the coil is then given by the product of the magnetic field and the coil's cross-sectional area. The diameter of the solenoid is 6.35 cm = 0.0635 m, so the cross-sectional area (A) is π*(diameter/2)2 = π*(0.0635 m / 2)2 = 0.00316 m2.

Therefore, the magnetic flux is Φ = BA = 0.015 T * 0.00316 m2 = 0.0000474 T*m2 = 0.0000474 Weber.

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